Limiting Sum including Stirling numbers

We get for the sum term

$$\sum_{k=1}^{n-1} {n\choose k} {n\brace n-k} (n-k)! \times k x^k \\ = n \sum_{k=1}^n {n-1\choose k-1} {n\brace n-k} (n-k)! \times x^k \\ = n x \sum_{k=0}^{n-1} {n-1\choose k} {n\brace n-1-k} (n-1-k)! \times x^k \\ = n x \times n! [z^n] \sum_{k=0}^{n-1} {n-1\choose k} (\exp(z)-1)^{n-1-k} x^k \\ = n x \times n! [z^n] (\exp(z)-1+x)^{n-1}.$$

We find with $x=1$

$$n \times n! \times [z^n] \exp((n-1)z) = n \times (n-1)^n.$$

We obtain for $x=(n-2)/(n-1)$

$$n \times \frac{n-2}{n-1} \times n! [z^n] (\exp(z)-1/(n-1))^{n-1}.$$

Now the coefficient extractor is

$$\frac{1}{2\pi i} \int_{|z|=1} \frac{\exp((n-1)z)}{z^{n+1}} (1-\exp(-z)/(n-1))^{n-1} \; dz.$$

With $n$ large this becomes

$$\frac{1}{2\pi i} \int_{|z|=1} \frac{\exp((n-1)z)}{z^{n+1}} \exp(-\exp(-z)) \; dz.$$

We now apply the Saddle point Algorithm for Cauchy Coefficient Integrals from pages 548 and 553 of Analytic Combinatorics by Flajolet and Sedgewick. (Theorem VIII.3) Our function $f(z)$ is

$$f(z) = (n-1)z-(n+1)\log z -\exp(-z)$$

and

$$f'(z) = (n-1) - (n+1)\frac{1}{z} + \exp(-z)$$

as well as

$$f''(z) = (n+1)\frac{1}{z^2} - \exp(-z).$$

Now let $\zeta_n$ be the positive real root of the saddle point equation $f'(\zeta_n) = 0$ so that

$$\zeta_n (n-1) - (n+1) + \zeta_n \exp(-\zeta_n) = 0$$

and

$$\zeta_n = \frac{n+1}{n-1+\exp(-\zeta_n)}.$$

This implies

$$\frac{n+1}{n} \lt \zeta_n \lt \frac{n+1}{n-1}$$

so that with $n$ large $\zeta_n \approx 1.$ We get for the asymptotics from equation 19 on the cited page

$$\frac{\exp(f(\zeta_n))}{\sqrt{2\pi f''(\zeta_n)}}$$

Now $f''(\zeta_n) = (n+1)/\zeta_n^2+(n-1)-(n+1)/\zeta_n$ and this evaluates to

$$\frac{\exp(n-1)\exp(-\exp(-1))}{\sqrt{2\pi (n-1)}}$$

Collecting everything from the two pieces we find for the first piece

$$\frac{1}{n^{n+1}} \times n\times (n-1)^n = \left(1-\frac{1}{n}\right)^n \sim \exp(-1)$$

and for the second piece

$$\frac{1}{n^{n+1}} \times n \frac{n-2}{n-1} n! \times \exp(-1) \frac{\exp(n)\exp(-\exp(-1))}{\sqrt{2\pi (n-1)}} \\ = \left(1-\frac{1}{n-1}\right) \exp(-1) \exp(-\exp(-1)) \frac{n!}{n^n} \frac{\exp(n)}{\sqrt{2\pi (n-1)}} \\ \sim \left(1-\frac{1}{n-1}\right) \exp(-1) \exp(-\exp(-1)) \frac{n!}{n^n} \frac{\exp(n)}{\sqrt{2\pi n}}.$$

We apply Stirling's formula to conclude and get

$$\left(1-\frac{1}{n-1}\right) \exp(-1) \exp(-\exp(-1)) \sim \exp(-1) \exp(-\exp(-1)).$$

We join the two pieces to obtain

$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{e}\left(1-e^{-1/e}\right)}$$

as proposed.

A different approach, first do $k=n-k$ so make the summation go in the inverse way and distribute, we get $$ \frac{1}{n^{n+1}}\sum_{k=1}^{n-1}\binom{n}{n-k}S\left( n,n-k\right) \left( n-k\right) !k\left( 1-\left( \frac{n-2}{n-1}\right) ^{k}\right)=\frac{1}{n^{n+1}}\left ( \sum_{k=1}^{n-1}\binom{n}{k}S\left( n,k\right) \left( k\right) !(n-k)\right )-\frac{1}{n^{n+1}}\left ( \sum_{k=1}^{n-1}\binom{n}{k}S\left( n,k\right) \left( k\right) !(n-k) \left (1-\frac{1}{n-1}\right )^{n-k}\right ), $$ Recall that $x^n=\sum _{k=1}^nS(n,k)x^{\underline{k}}=\sum _{k=1}^nS(n,k)\binom{x}{k}k!,$ so that the first part becomes $$\frac{1}{n^{n+1}}(n^{n+1}-n(n^n-(n-1)^n)),$$ where the negative part comes from, for example, OEIS. the limit of this is $1/e.$

Now, for the second part, one has $$\frac{1}{n^{n+1}}\left ( \sum_{k=1}^{n-1}\binom{n}{k}S\left( n,k\right) \left( k\right) !(n-k) \left (1-\frac{1}{n-1}\right )^{n-k}\right )=\frac{1}{n^{n+1}}\left ( \sum_{k=1}^{n-1}\binom{n}{n-k}S\left( n,k\right) \left( k\right) !(n-k) \sum _{l=0}^{\infty}\binom{n-k}{l}(-1/(n-1))^l\right ),$$ exchanging the two sums and using $\binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c},$ one gets $$\frac{1}{n^{n+1}}\sum _{l=0}^{\infty}\binom{n}{l}(-1/(n-1))^l\sum _{k=1}^{n-1}\binom{n-l}{n-k-l}k!S(n,k)(n-k),$$ which looks a lot like the first problem. Except that now $x=n-l.$ So we can check that the inner sum is $$n(n-l)^n-n((n-l)^n-(n-l-1)^n),$$ so if one pushes the limit inside (ALERT: I am used to combinatorics, probably one has to check something about that pushing the limit) one then can see using that $\binom{n}{l}\sim n^l/l!,$ that one gets $$-\sum _{l=0}^{\infty}\frac{(-1)^{l}}{l!}e^{-(l+1)}=\frac{-1}{e}e^{-e^{-1}}.$$ Which agrees with your computation.